## [SOLVED] (not) About 'Almost locked sets'.

Everything about Sudoku that doesn't fit in one of the other sections

### [SOLVED] (not) About 'Almost locked sets'.

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EDIT
My synopsis if you want to 'cut to the chase'.
Here...
Hidden Text: Show
January 22 2015
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Method
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Choose three squares A,B and X anywhere on the Sudoku grid.
Squares A,B and X can have any number of candidates.
Square A and square B must share at least one common candidate.

Choose a common candidate.
Perform a test with the chosen common candidate in square A set to 'off' and note the result obtained in square X.
Perform a test with the chosen common candidate in square B set to 'off' and note the result obtained in square X.

If the two results are not incompatible then the tests are inconclusive.
Choose a different candidate or different squares and try again.

If the two results are incompatible...
It's telling us that the puzzle won't complete with square A and/or square B set with their common factor in these states.

It's telling us that to complete the puzzle we would need to choose one of these three options.

i Change the state of the chosen common candidate in square A from 'off' to 'on'.

ii Change the state of the chosen common candidate in square B from 'off' to 'on'.

iii Change the state of the chosen common candidate in both square A and square B from 'off' to 'on'.
This third option is not applicable if square A and square B are in the same row or column or box.

What we don't know...
We don't know which option to choose.

What we do know...
We know that whichever option we chose...
At least one of the two squares A and B would then contain the chosen common candidate.

What's the point?
Any square on the grid that can see both square A and square B may have the chosen common candidate eliminated.

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Examples of incompatible results with a chosen common candidate 4
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X=12345678
"The value of square X is 7 when A is not 4. The value of square X is 8 when B is not 4"

X=12345678
"The value of square X is 7 when A is not 4. The value of square X is not 7 when B is not 4"

X=12345678
"The value of square X is 1 or 2 when A is not 4. The value of square X is 3 or 4 when B is not 4"

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Real world examples
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Code: Select all
`*--------------------------------------------------------------------* |*18     7      6      | 2      13     9      | 4      5     *38     | |-12458  3      14589  |X47     1457  A14     |*78     2789   6      | | 245    259    459    | 467    34567  8      | 37     279    1      | |----------------------+----------------------+----------------------| | 1458   1568   1458   | 3      9      46     | 1578   4678   2      | | 3      568    2      | 1      468    7      | 9      468    458    | | 7      1689   1489   | 468    2      5      | 138    468    348    | |----------------------+----------------------+----------------------| | 6      1258   1358   | 9      13478  1234   | 578    478    4578   | | 9      258    358    | 478    3478   234    | 6      1      4578   | |B18     4      7      | 5      168   -16     | 2      3      9      | *--------------------------------------------------------------------*`
A=r2c6, B=r9c1, X=r2c4
A=4 => X=r2c4 = 7
B=8 => r1c9=8 -> r2c7=7 -> X=r2c4 = 4
=> A or B is 1, r2c1,r9c6<>1

Code: Select all
`.------------------.------------------.-------------.| 2467   24   5    | 3    2467   8    | 679  1  679 || 2367   29   2369 | 127  1267   126  | 8    5  4   || 467    1    8    | 5    9      46   | 3    2  67  |:------------------+------------------+-------------:| 45     3    7    | 8    124    124  | 59   6  159 || 9      45   X16  | A17  13467  1346 | 5-7  8  2   || *126   8    *126 | 9    16-7   5    | 4    3  B17 |:------------------+------------------+-------------:| 13    6    13    | 4    5      9    | 2    7  8   || 8     7    4     | 12   123    123  | 56   9  56  || 25    259  29    | 6    8      7    | 1    4  3   |.------------------.------------------.-------------.`
A=r5c4, B=r6c9,X=r5c3
A=1->X=6
B=1->X=1
=> A or B is 7, r5c7,r6c5<>7

Code: Select all
`.-----------------.------------------.------------------.| 9      13   2   | 5     6    38    | 48    7     1348 || 4      8    67  | 137   127  237   | 9     2356  35   || 167    137  5   | 9     4    2378  | 268   236   138  |:-----------------+------------------+------------------:| 5678   4    678 | 68    279  25679 | 1     235   358  || 2      9    178 | 148   3    B57   | 478   *45    6   || 15678  57   3   | 1468  127  2567  | 2478  9     458  |:-----------------+------------------+------------------:| 78     2    48  | A67   5    1     | 3     X46    9   || 157    157  147 | 367   79   369-7 | 46    8     2    || 3      6    9   | 2     8    4     | 5     1     7    |'-----------------'------------------'------------------'`
A=r7c4, B=r5c6, X=r7c8
A=6->X=4
B=5->X=6
=> A or B is 7, r8c6<>7

Code: Select all
`.---------------.-------------.-------------.| 5   4    3    | 7    2  9   | 1    6  8   || 2   1    b69  | 4    8  a36 | 9-5  7  B35 || 68  89   7    | 1    5  36  | 49   2  34  |:---------------+-------------+-------------:| 7   59   c29  | 259  1  4   | 3   8  6    || 68  589  469  | 59   3  7   | 2   1  45   || 1   3    X24  | 25   6  8   | A45  9  7   |:---------------+-------------+-------------:| 9   7    8    | 3    4  2   | 6   5  1    || 4   2    5    | 6    7  1   | 8   3  9    || 3   6    1    | 8    9  5   | 7   4  2    |'---------------'-------------'-------------'`
A=r6c7, B=r2c9, X=r6c3
A=4->X=2
B=3->X=4
=> A or B is 5, r2c7<>5

Code: Select all
`*--------------------------------------------------* | 5    4    3    | 7    2    9    | 1    6    8    | | 2    1   *69   | 4    8    36   | 59   7    35   | | 68   89   7    | 1    5    36   | B49   2   3-4  | *----------------+----------------+----------------| | 7    59   29   | 259  1    4    | 3    8    6    | | 68   589 X469  | 59   3    7    | 2    1   A45   | | 1    3    24   | 25   6    8    | 5-4  9    7    | *----------------+----------------+----------------| | 9    7    8    | 3    4    2    | 6    5    1    | | 4    2    5    | 6    7    1    | 8    3    9    | | 3    6    1    | 8    9    5    | 7    4    2    | *--------------------------------------------------*`
A=r5c9, B=r3c7, X=r5c3
A=5->X=4
B=9=>r2c3=9->X=6
=> A or B is 4, r3c9,r6c7<>4

Code: Select all
`.------------------.---------------.------------------.| 8    3     9     | 45-6 2  X45   | 156   B16    7   || 1    6     2     | a358 7  b358  | 359   389    4   || 45   47    457   | A368 9  1     | 3-6   238-6  23  |:------------------+---------------+------------------:| 49   2     134   | 149  8  6     | 1349  7      5   || 7    1489  1348  | 149  5  249   | 1349  1239   6   || 469  5     146   | 7    3  e249  | 8     c129   d29 |:------------------+---------------+-------- ---------:| 569  1789  15678 | 2    4  35789 | 3679  369    39  || 3    4789  478   | 89   6  789   | 2     5      1   || 2    79    567   | 359  1  3579  | 3679  4      8   |'------------------'---------------'------------------'`
A=r3c4, B=r1c8, X=r1c6
A<>6=>r3c4,r2c46=358->X=4
B<>6=>r6c89=29=>r6c6=4->X=5
=> A or B is 6, r1c4,r3c7,r3c8<>6

Code: Select all
`.---------------.----------.------------.| 3  5      2   | 4  7   1 | 6   9    8 || 7  89     89  | 6  3   5 | 14  14   2 || 6  4      1   | 2  9   8 | 5   3    7 |:---------------+----------+------------:| 5  1      3   | 7  4   2 | 9   8    6 || 8  X27    47  | 5  6   9 | 3   24   1 || 9  A26    46  | 1  8   3 | 24  7    5 |:---------------+----------+------------:| 2  d79-6  679 | 8  15  4 | 17  B56  3 || 4  c78    5   | 3  12  6 | 78  12   9 || 1  3      b68 | 9  25  7 | 28  a56  4 |'---------------'----------'------------'`
A=r6c2, B=r7c8, X=r5c2
A<>6->X=7
B<>6=>r9c8=6=>r9c3=8=>r8c2=7->X=2
=> A or B is 6, r7c2<>6

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Hi
I've started a new thread because I don't want to clutter Instinct's journal.
Here ---> http://forum.enjoysudoku.com/my-sudoku-learning-journal-t32233.html

Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is the same, then something is known about X because (at least) one of A or B must be not 4.
Square X can now be assigned a value or at least have something eliminated from it.
(I think this is known as a 'Kraken' method)
So far, so good.

Now with reference to Instinct's post about 'Almost locked sets'...

Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

Or have I over-presumed.
Last edited by bat999 on Thu Jan 29, 2015 5:24 pm, edited 7 times in total.
bat999
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Location: UK

### Re: About 'Almost locked sets'.

Hi bat, your descriptions are hard to follow as they are unusual by the standards of this (or any other reputable) forum but I'll try to answer them as well as I can.

What you have do is ensure that you have covered 100% of possible cases, with a common outcome in your object cell X. If this common outcome is, say, y then you can assign y in X.

What you have to consider in your 2 bi-value squares, is:

(1) A is 4 and B is 4, (2) A is not 4 and B is 4, (3) A is 4 and B is not 4 & (4) A is not 4 and B is not 4 (unless they are the only 2 4's in a row, column or box.)

Since you say that A and B are in the same row, column or box then (1) can't happen and can be ignored. However it looks like you have only considered cases (2) and (3). To correctly apply the method you also have to consider case (4) ie both A and B are not 4 (unless they are the only 2 4's in a row, column or box,which they aren't in instinct's example.)

You have made a similar mistake in the second section.

Your statement: If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect. So one of the squares A or B must be 4. Is not quite right.

What you can conclude is : If the result in X is the not the same, then nothing is known about X, but it indicates that at least one of the values chosen for A or B were incorrect.

So you can't conclude that one of the squares must be 4. They might both be not 4 (unless of course they are the only 2 4's in a row, column or box, which they aren't in instinct's example.)

Now let's have a look at Instinct's Diagram B. What he says here is :

1. r2c6 <> 4 = 1;

2. r2c4 <> 4 = 7 => r2c7 <> 7 = 8 => r1c9 <> 8 = 3 => r1c5 <> 8 = 1; Contradiction (2 1's in Box 2) => at least one of r2c46 must be 4.

This is logically correct, but most experienced solvers regard proof by contradiction as inelegant, if a constructive proof of similar complexity is available.

Fortunately this is true in this case - simply by rearranging things a bit.

A constructive proof is : r2c4 <> 4 = 7 => r2c7 <> 7 = 8 => r1c9 <> 8 = 3 => r1c5 <> 3 = 1 => r2c6 <> 1 = 4.

A more standard notation for this chain is : (4=7) r2c4 - (7=8) r2c7 - (8=3) r1c9 - (3=1) r1c5 - (1=4) r2c6.

If you read the chain from left to right you assume that r2c4 <> 4 and end up concluding that r2c6 = 4.

If you read the chain from right to left you assume that r2c6 <> 4 and end up concluding that r2c4 = 4.

You are then entitled to conclude that at least one of r2c46 must be 4 (They might both be 4 but they can't both be <> 4.) and you have done so without producing a contradiction.

This type of chain is called an XY chain (each of the cells in the chain has 2 digits X and Y with Y in each cell being the same as X in the next cell in the chain, with each cell in the chain sharing a row, column or box with the proceeding cell.

Notice that the "contradiction" part of instinct's argument has been replaced by a linking digit in my XY chain (which says that if r1c5 = 1 then r2c6 <> 1 when read from left to right, or if r2c6 = 1 then r1c5 <> 1 when read from right to left).

Hope this helps.

Leren
Leren

Posts: 3606
Joined: 03 June 2012

### Re: About 'Almost locked sets'.

Leren wrote:(4) A is not 4 and B is not 4 (unless they are the only 2 4's in a row, column or box.)
Hi
This is not a mistake.
It doesn't matter how many squares in the same row or column or box have the 4 as a candidate.
If you can find any two squares in the same row or column or box that give the same result in the object square when 'off', that's it.
It's not necessary to test any more squares.

It's not clear from the rest of your post whether the answer to my question is 'yes'.
bat999
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Location: UK

### Re: About 'Almost locked sets'.

bat 999 wrote:

It doesn't matter how many squares in the same row or column or box have the 4 as a candidate.
If you can find any two squares in the same row or column or box that give the same result in the object square when 'off', that's it.
It's not necessary to test any more squares.

If that's what you continue to believe then good luck !

Here is a recent Kraken move of mine

Code: Select all
`*---------------------------------------------------------------------------------*| g9-6dG   249     26       | 49C     7       5        | 3       8       1        ||  5       19      3        | 6       8       19B      | 7       4       2        ||  8       14      7        | 134     24      123      | 9       6       5        ||---------------------------+--------------------------+--------------------------||f 679     5       4        | 2       69      379      | 1       39      8        || a167cF  e39      8        | 147aD   5       17A      | 2       39      46bE     ||fb169    d239     26       |c1349    469     8        | 5       7       46       ||---------------------------+--------------------------+--------------------------||  4       8       1        | 79      29      279      | 6       5       3        ||  3       7       5        | 8       1       6        | 4       2       9        ||  2       6       9        | 5       3       4        | 8       1       7        |*---------------------------------------------------------------------------------*Kraken Row 5 Digit 1 :1 r5c1 - r6c1 = (1-3) r6c4 = r6c2 - (3=9) r5c2 - r46c1 = r1c1 - 6 r1c1;1 r5c4 - 4 r5c4 = (4-6) r5c9 = r5c1                           - 6 r1c1;1 r5c6 - (1=9) r2c6 - (9=4) r1c4 - r5c4 = (4-6) r5c9 = r5c1   - 6 r1c1; => - 6 r1c1; stte`

What you seem to be saying is that was only necessary for me to consider any 2 of the 3 1's in Row 5 to show that r1c1 <> 6 instead of all three 1's.

In cases like this I intend to stick to considering all three 1's.

Leren
Leren

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Joined: 03 June 2012

### Re: About 'Almost locked sets'.

bat999 wrote:It's not clear from the rest of your post whether the answer to my question is 'yes'.

no
eleven

Posts: 2106
Joined: 10 February 2008

### Re: About 'Almost locked sets'.

eleven wrote:no

OK, thanks.
bat999
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Location: UK

### Re: About 'Almost locked sets'.

Bat999,
Your first section is correct, only this is NOT the Kraken method, which starts with a WEAK link, in other words a positive assumption rather than the negative "if NOT" of a strong link. I think Leren is describing the conditions necessary for a Kraken. Your initial assumption is that either (NOT one 4) OR (NOT another 4 in same sector) must be true, and this is correct regardless of whether the 4s involved are in bivalue cells or any other cells. E. G. the 2 cells could just as well be 12345 and 234567. If you get the same result for both one 'NOT 4' and the other 'NOT 4', then you CAN deduce that the result is true, no matter what the result may be. That is simple and clear logic. (Bear in mind that very often a wrong assumption can lead to the correct answer for a cell.)

Your second section is incorrect, for the reasons that Leren describes in detail.
Last edited by gurth on Sat Jan 10, 2015 6:53 am, edited 1 time in total.
gurth

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Joined: 11 February 2006
Location: Cape Town, South Africa

### Re: About 'Almost locked sets'.

Leren wrote:This is logically correct, but most experienced solvers regard proof by contradiction as inelegant, if a constructive proof of similar complexity is available.

Leren

This statement is correct logically and any otherwise you can think of, only it is an understatement, most experienced solvers seem to have a phobia about contradictions. Which is absolutely absurd. There is nothing inelegant about contradictions. "Inelegant" is a purely subjective judgement, containing no logic whatsoever. Perhaps these people who hate contradictions so much, have a low tolerance for confrontations and disharmony of any sort.

I will be the first to concede that a proof, for any given case, that is simpler and/or shorter than a contradiction, is more elegant. But not if it is the same length and equally complex.

.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

### Re: About 'Almost locked sets'.

Hi bat999,

bat999 wrote:Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

Yes it's true, as long as you're careful amout what is meant by "the result in X is the not the same".
It might be better to think of it as that "the results in X are incompatible".

That would cover cases where:
1) one option forces some value into X, and the other option forces a different value into X.
2) one option forces X to be 'd', and the other eliminates 'd' from X.
3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options.

(If X is a bivalue cell, then the 3 cases are more or less equivalent).

Note: the logic here, has nothing to do with Almost Locked Sets or that A and B are bi-value cells, and it doesn't require that cells A and B can see each other.
blue

Posts: 874
Joined: 11 March 2013

### Re: About 'Almost locked sets'.

@ gurth
The term 'kraken' was inappropriate for the first part of my question.

@ blue
Now I seem to have two opposite answers to my question.
bat999
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Location: UK

### Re: About 'Almost locked sets'.

Sorry, i had to leave urgently yesterday.
But Gurth and blue already clarified it.
If by "result in X not the same" you mean different numbers, it is true, if you mean different candidates, then not. As long as a common candidate is left in X, both values can be correct.
eleven

Posts: 2106
Joined: 10 February 2008

### Re: About 'Almost locked sets'.

@ eleven

gurth said no and blue said yes.

This I don't understand:-
If by "result in X not the same" you mean different numbers, it is true, if you mean different candidates, then not.
As long as a common candidate is left in X, both values can be correct.

I will search the www for an explanation and examples.
bat999
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Location: UK

### Re: About 'Almost locked sets'.

What i mean is simply:
Suppose X has candidates 123, and A=4 reduces it to 12, while B=4 reduces it to 23, the "result" for the candidates in X is different, but still both are possible, because X could resolve to 2.
eleven

Posts: 2106
Joined: 10 February 2008

### Re: About 'Almost locked sets'.

eleven wrote:... the "result" for the candidates in X is different, but still both are possible, because X could resolve...

I've tightened up the question.
"Different" isn't strong enough.

Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is contradictory, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

PS
I've just read blue's post again.
"It might be better to think of it as that "the results in X are incompatible".
I think me and blue are on the same wavelength.
--------------------------------------------------
These are my two examples of 'contradictory' results.
"The value of square X is 7 when A is not 4. The value of square X is 8 when B is not 4"
and
"The value of square X is 7 when A is not 4. The value of square X is not 7 when B is not 4"
--------------------------------------------------
And this was how blue expressed it...
1) one option forces some value into X, and the other option forces a different value into X.
2) one option forces X to be 'd', and the other eliminates 'd' from X.
But I haven't thought this one through yet.
3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options.
bat999
2017 Supporter

Posts: 677
Joined: 15 September 2014
Location: UK

### Re: About 'Almost locked sets'.

yes
eleven

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